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$A$ is symmetric positive definite and $A = LDL^T$, where $L$ is unit lower triangular and $D$ is diagonal. I want to prove that the main-diagonal entries of $D$ are all positive.

I have tried $\det(A)>0 \Rightarrow \det(LDL^T)>0$. Since $\det(L)=\det(L^T)=1$, $\det(D)$ must be positive. But, that doesn't mean all of the main diagonal entries of D are positive. Should I be using a different property?

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If $X$ is non-singular and $A$ is PD then $X^T A X$ is PD. Take $X = L^{-1}$. –  user2468 Feb 25 '12 at 20:16
    
@J.D. Do you mean let $X=(L^{-1})^T$? So that $X^TAX=D$ –  Ashley Feb 27 '12 at 3:46
    
Yes. That's what I meant. –  user2468 Feb 27 '12 at 4:12

2 Answers 2

There are probably many ways of doing it. Here is one:

Note first that $A$ admits a square root, i.e. there is a symmetric positive definite matrix $A^{1/2}$ such that $(A^{1/2})^2=A$. This is done by writing $A$, via the Spectral Theorem, as $$ A=\sum_{j=1}^n \lambda_j P_j, $$ where $\lambda_1,\ldots,\lambda_n$ are the eigenvalues of $A$ counting multiplicities, and $P_1,\ldots,P_n$ are pairwise orthogonal projections of rank one (i.e. the projections onto the corresponding eigenspaces). Then define $$ A^{1/2}=\sum_{j=1}^n \lambda_j^{1/2} P_j. $$

Now, since $L$ is invertible, we can write $$ D=MAM^T, $$ where $M=L^{-1}$. Considering the canonical basis $\{e_1,\ldots,e_n\}$, we have $$ D_{kk}=\langle De_k,e_k\rangle = \langle MAM^Te_k,e_k\rangle =\langle A^{1/2}M^Te_k, A^{1/2}M^Te_k\rangle\geq0. $$ But $D$ is invertible, so $D_{kk}\ne0$, and so $D_{kk}>0$ for all $k=1.\ldots,n$.

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Let $e_j$ the $j$-th vector of the canonical basis. Since $L$ is invertible we are allowed to write $(L^t)^{-1}e_j$ and since $A$ is positive definite $$0<((L^{-1})^te_j)^tA(L^{-1})^te_j=e_j^tL^{-1}A(L^{-1})^{t}e_j=e_j^tDe_j=D_{jj}$$ and we are done.

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